如何通过以下操作手动跟踪链接列表 X？

Question

我正在尝试在下面回答这个问题，但我被卡住了。

通过以下操作手工追踪一个链表X：

X.add("Fast"); 
X.add("Boy"); 
X.add("Doctor"); 
X.add("Event"); 
X.add("City"); 
X.addLast("Zoo"); 
X.addFirst("Apple"); 
X.add (1, "Array");
X.remove("Fast");
X.remove (2);
X.removeFirst ();
X.removeLast ();

这就是我手动追踪它的方式，我试图了解我是否/在哪里做错了，因为我在其他地方看到了令人困惑的答案：

快

男孩快

医生男孩快

事件医生男孩快

城市活动医生男孩快

动物园城市活动医生男孩快

动物园城市活动医生男孩快速苹果

动物园城市活动医生男孩快速数组苹果

动物园城市活动医生男孩数组苹果

动物园城市事件医生阵列苹果

动物园城市事件医生数组

城市事件医生数组

Answer 1

执行操作后会得到如下列表：

您的 LinkedList 从头部开始。现在首先，你必须从头部插入，否则到最后。

快

男孩->快

医生->男孩->快

事件->医生->男孩->快

城市->事件->医生->男孩->快速

City->Event->Doctor->Boy->Fast->Zoo // 错误 Zoo->City->Event->Doctor->Boy->Fast

Apple->City->Event->Doctor->Boy->Fast->Zoo // 错误 Zoo->City->Event-> Doctor->Boy->Fast->Apple

Apple->Array->City->Event->Doctor->Boy->Fast->Zoo // 错误，因为链表应该从苹果开始，但它不完全正确，因为 LinkedList 中没有索引它应该在你身上。

同样，您可以执行其他操作。

Answer 2

X.add("Fast"); 
X.add("Boy"); 
X.add("Doctor"); 
X.add("Event"); 
X.add("City"); 
City->Event->Doctor->Boy->Fast
tail    ----- head
top   ---- bottom
top........0

X.addLast("Zoo"); 
X.addFirst("Apple");
Zoo->City->Event->Doctor->Boy->Fast->Apple

X.add (1, "Array");
Zoo->City->Event->Doctor->Boy->Fast->Array->Apple

X.remove("Fast");
Zoo->City->Event->Doctor->Boy->Array->Apple

X.remove (2);
Zoo->City->Event->Doctor->Array->Apple

X.removeFirst ();
Zoo->City->Event->Doctor->Array

X.removeLast ();
City->Event->Doctor->Array

Answer depends on , what each operation means,
Here is how I perceived them:
add("A") -> will append "A" as head to the linked list.
addFirst("A") -> will add "A" as head to the linked list i.e. first element.
addLast("A") -> will append "A" as tail to the linked list.
add (i, "A") -> will add "A" at i th position from head(i=0,1,2,..)
remove("A") ->  will remove "A" from LL
remove(i) -> will remove element from i th position
removeFirst() -> will remove (head)first element from LL.
removeLast() -> will remove (tail)last element from LL.
It will be helpful if you can provide this sort of detail.